Calculate the pH at which Mg(OH)2 begins to precipitate from a solution containing 0.1 M Mg2+ ions? Ksp for Mg(OH)2 = 1.0 x 10−11.

1 Answer
Stefan V.
Jan 31, 2016

pH=9

Explanation:

The idea here is that you need to use magnesium hydroxide's solubility product constant to determine what concentration of hydroxide anions, OH−, would cause the solid to precipitate out of solution.

As you know, the dissociation equilibrium for magnesium hydroxide looks like this

Mg(OH)2(s]⇌Mg2+(aq]+2OH−(aq]

The solubility product constant, Ksp, will be equal to

Ksp=[Mg2+]⋅[OH−]2

Rearrange to find the concentration of the hydroxide anions

[OH−]= ⎷Ksp[Mg2+]

Plug in your values to get

[OH−]=√1.0⋅10−110.1=√1.0⋅10−10=10−5M

As you know, you can use the concentration of hydroxide anions to find the solution's pOH

pOH=−log([OH]−)

pOH=−log(10−5)=5

Finally, use the relationship that exists between pOH and pH at room temperature

pOH + pH=14

to find the pH of the solution

pH=14−5=9

So, for pH values that are below 9, the solution will be unsaturated. Once the pH of the solution becomes equal to 9, the solution becomes saturated and the magnesium hydroxide starts to precipitate.

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